**“****Area and its Boundary****” **is **Chapter 11** from **Merry Math V **for students of** Class 5th **of **JKBOSE**. This post is about** Area and its Boundary** **Class 5 JKBOSE Solutions****. **You read about **Tenths and Hundredths Class 5 JKBOSE Solutions** in a previous post. Let’s get started:

**Area and its Boundary Class 5 JKBOSE Solutions**

**Page No. 161 Solutions**

**Whose Slice is Bigger?**

Salim and Rukaiya bought *aam paapad *[dried mango slice] from a shop.

Their pieces looked like these.

Both could not make out whose piece was bigger. Suggest some ways to find out whose piece is bigger. Discuss.

A friend of Salim and Rukaija showed one way, using small squares.

The length of piece A is 6 cm.

So, 6 squares of side 1 cm can be arranged along its length.

The width of piece A is 5 cm.

So, 5 squares can be arranged along with their width.

**Page No. 162 Solutions**

**Altogether how many squares can be arranged on it? _________**

Ans. The length of piece A is 6 cm. The width of piece A is 5 cm. So, the number of squares of side 1 cm can be altogether arranged on it = (6 × 5) = 30 squares.

**So, the area of piece A = _________ square cm.**

Ans. So the area of piece A = **30 **square cm.

**In the same way, find the area of piece B.**

Ans. In the same manner, 3 squares of side 1 cm can be arranged along its width and 11 squares of side 1 cm can be arranged along its length. So, the total number of squares = (3 × 11) = 33 squares.

The area of piece B = 33 square cm.

**Who had the bigger piece? How much bigger?**

Ans Rukaiya had the bigger piece because the area of her piece was bigger. It is bigger by 3 square cm than piece A.

**Cover with Stamps**

This stamp has an area of 4 square cm. Guess how many stamps will cover this big rectangle.

Ans. I think 24 stamps will cover this big rectangle.

**Page No. 163 Solutions**

**Check your guess **

**a. Measure the yellow rectangle. It is _______ cm long.**

Ans. 12 cm

**b. How many stamps can be placed along its length? _________**

Ans. 6 stamps.

**c. How wide is the rectangle? _______ cm**

Ans. 8 cm.

**d. How many stamps can be placed along its width? ________**

Ans. 4 stamps.

**e. How many stamps are needed to cover the rectangle? _______**

Ans. 6 x 4 = 24 stamps.

**f. How close was your earlier guess? Discuss.**

Ans. My guess was exactly right.

**g. What is the area of the rectangle? _______ square cm.**

Ans. Area of rectangle = 24 x 4 = 96 square cm.

**h. What is the perimeter of the rectangle? _______ cm.**

Ans. The perimeter of the rectangle = 2 (12 + 8) = 40 cm.

**Practice Time **

**a. Adnan plans to tile his kitchen floor with green square tiles. Each side of the tile is 10 cm. His kitchen is 220 cm in length and 180 cm wide. How many tiles will he need?**

Sol. Length of the kitchen = 220 cm.

Width of the kitchen = 180 cm

Area of kitchen = L x B = (220 × 180) square cm = 39600 square cm.

Length of a tile = 10 cm

Width of a tile = 10 cm

Area of a tile = (10 × 10) square cm = 100 square cm.

Number of tiles = Area of the floor/Area of a tile.

= 39600/100 = 396 tiles.

**b. The fencing of a square garden is 20 m in length. How long is one side of the garden?**

Ans. The length of fencing of a square park = 20m

The length of fencing = Perimeter of the park.

The perimeter of a park = 4 × side.

4 × side = 20 m

Side = 20/4 = 5 m.

**c. A thin wire 20 cm long is formed into a rectangle. If the width of this rectangle is 4 cm, what is its length?**

Ans. Length of wire = 20 cm.

The perimeter of the rectangle = Length of wire = 20 cm.

Let the length of the side of the rectangle = x cm

Width of rectangle = 4 cm.

Perimeter of rectangle = 2 (length + width) = 2 (x + 4)

20 cm = 2x + 8

Or 2x + 8 = 20

2x = 20 – 8

2x = 12

x = 12/2 = 6

Length of the rectangle = 6 cm.

**Page No. 164 Solutions**

**d. A square carom board has a perimeter of 320 cm. How much is its area?**

Ans. Given perimeter of the carrom board is 320 cm

We know that the perimeter of the square = 4 × side

Side of a carrom board = 320/4 = 80 cm.

Area of a square = side × side.

= 80 × 80 = 6400 square cm.

**e. How many tiles like the triangle given here will fit in the white design?**

Ans. Area of the design = 1 full square + 4 triangles.

(2 Triangles = 1 full square).

So, the area of the design = 3 full squares = 3 square cm.

Number of tiles = Area of design/Area of 1 tile

= 3 × 2/1 = 6.

**f. Ambreen, Ulfat, Mudasir and Kabir made greeting cards. Complete the table for their cards:**

Ans. i) Length of Ambreen’s card = 10 cm.

Width of Ambreen’s card = 8 cm.

Perimeter of Ambreen’s card = 2 (length + width)

= 2 (10 + 8) = 36 cm.

Area of Ambreen’s card = Length × Width.

= 10 × 8 square cm = 80 square cm.

ii) Length of Mudasir’s card = 11 cm.

Width of Mudasir’s card = ?

Perimeter of Mudasir’s card = 44 cm.

Also, Perimeter = 2 (L + W)

44 = 2 (11 + W)

44 = 22 + 2W

22 + 2W = 44

2W = 44 – 22

2W = 22

W = 22/2 = 11 cm.

Area of Mudasir’s card = Length × Width.

= 11 cm × 11 cm = 121 square cm.

iii) Width of Ulfat’s card = 8 cm.

The area of Ulfat’s card = 80 square cm.

Also, area = Length × Width.

80 square cm = Length × 8 cm

Length = 80/8 cm = 10 cm.

The perimeter of Ulfat’s card = 10 cm × 8 cm = 80 square cm.

iv) Perimeter of Kabir’s greeting card = 40 cm.

The area of Kabir’s greeting card = 100 square cm.

We know that Perimeter = 2 (Length + Width)

= 40 cm = 2 (Length + Width)

= 40/2 cm = Length + Width

= 20 cm – Length = Width………. (i)

In the second case.

Area of card = 100 square cm.

Area = Length × Width

100 = Length × 20 cm – Length

100 = 20Length – Length^{2}

20Length – Length^{2 }= 100

Length^{2} – 20Length + 100 = 0

Length^{2 }– 10Length – 10 Length – 100 = 0

Length (Length – 10) – 10 (Length – 10) = 0

Length (Length – 10) = 10 (Length – 10)

Length = 10 cm

Width = 20 – 10 = 10 cm.

**My Belt is the Longest!**

Take a thick paper sheet of length 14 cm and width 9 cm. You can also use an old postcard.

**What is its area? What is its perimeter?**

Ans. Length of the sheet of paper = 14 cm

Width of the sheet of paper = 9 cm.

Area of sheet of paper = 14 cm × 9 cm = 126 square cm.

The perimeter of the sheet of paper = 2 (Length + Breadth)

= 2(14 + 9) cm = 2 (23) = 46 cm.

**Now cut strips of equal sizes out of it. Using tape join the strips, end to end, to make a belt.**

Ans. Cut 9 strips of 1 cm width and 14 cm length from this thick sheet of paper and join them end to end with tape to make a belt.

**How long is your belt? _____**

Ans. Length of the belt 14 × 9 cm = 126 cm.

**What is its perimeter ____**

Ans. The perimeter of the belt will be 2(126 + 1) = 2 × 127 = 254 cm.

**Discuss**

**Why did some of your friends get longer belts than others?**

Ans. Some of my friends get longer belts than others as the width of their belts is shorter than that of the other’s belts.

**Is the area of your belt the same as the area of the postcard? Why or why not?**

Ans. Yes, the area of the belt is the same as the area of the postcard because it is made out of the postcard without any wastage.

**What will you do to get a longer belt next time?**

Ans. I will get a longer belt next time by ensuring that the width of my belt should be shorter than other’s belts.

**Page No. 166 Solutions**

**People People Everywhere**

A) You can play this game on the ground.

Make two squares of one square metre each.

Divide your class into two teams. Ready to play!

Try these in your teams –

**How many of you can sit in one square metre? _______**

Ans. Three of us can sit in one square metre.

**How many of you can stand in it?**

Ans. Four of us can stand in it.

**Which team could make children stand in their square? How many?**

Ans. Team B could make more children to stand in their square. They are five in number.

**Which team could make more children sit in their square? How many?**

Ans. Team B could make more children to sit in their square. They are four in number.

**Measure the length of the floor of your classroom in metres. Also, measure the width. **

**What is the area of the floor of your classroom in square metres? _____**

Ans. Suppose the length of the floor of our classroom is 11 metres and the width of the floor of our classroom is 9 metres.

The area of our classroom = Length × Width.

= 11 metres × 9 metres = 99 square metres.

**How many children are there in your class?**

Ans. There are 66 children in our class.

**So how many children can sit in one square metre?**

Number of children who can sit in one square metre = Total number of students / Total area of the floor

= 99/66 = 3/2 = 1.5

So, 1.5 students can sit in one square metre.

**Page No. 167-168 Solutions**

**Share the Land**

Mubeena is a farmer who wants to divide her land equally among her three children – Asmat, Iram and Altaf. She wants to divide the land so that each piece of land has one tree. Her land looks like this.

** Can you divide the land equally? Show how you will divide it. Remember each person has to get a tree. Colour each person’s piece of land differently.**

Ans. Yes, I can divide the land equally among Mubeena’s three children. Asmat, Iram and Altaf such that each gets one tree. Each person’s piece of land is coloured differently as shown in the figure.

**If each square on this page is equal to 1 square metre of land, how much land will each of her children get? ________ square m **

Ans. Total number of squares = 90.

Area of the land = 90 square metres.

Each child will get an equal share which is equal to 90/3 = 30 square metres

**Asmat, Iram and Altaf need wire to make a fence.**

**Who will need the longest wire for fencing? ________**

i) Asmat’s Land

The perimeter of Asmat’s land for fencing = (4 + 3 + 1 + 6 + 3 + 9) = 26 metres.

ii) Iram’s Land

The perimeter of Iram’s land for fencing = (3 + 3 + 3 + 1 + 3 + 1 + 3 + 4 + 6) = 27 metres.

iii) Altaf’s Land

The perimeter of Altaf’s land for fencing = (3 + 1 + 3 + 3 + 9 + 3 + 3 + 1) = 26 metres.

So, Iram will need a longer wire for fencing.

**How much wire in all will the tree need? _______**

Ans. Total wire needed by them = (26 + 26 + 27) = 79 metres.

**Practice Time **

A. Look at the table. If you were to write the area of each of these which columns, would you choose? Mark a (🗸)

Ans.

**Page No. 169 Solutions**

**B. Draw a square of 9 cm. Write A on it. Draw another square with double the side. Write B on it.**

Ans. Let us draw a square A of 9 square cm. The side of this square is 3 cm. Also, draw a square B of a side length equal to 6 cm.

**Answer these –**

**The perimeter of square A is _______ cm.**

Ans. The perimeter of square A is 3 cm × 4 = 12 cm.

**The side of square B is _______ cm.**

Ans. The side of square B is 3 cm × 2= 6 cm.

**The area of square B is _______ square cm.**

Ans. The area of square B is 6 cm × 6 cm = 36 square cm.

**The area of square B is _______ times the area of square A.**

Ans. The area of sq. B is four times of the area of square A.

**The perimeter of square B is _______ cm.**

Ans. The perimeter of square B is 6 cm × 4 = 24 cm.

**The perimeter of square B is _______ times the perimeter of square A.**

Ans. ** **The perimeter of square B is two times the perimeter of square A.

**Page No. 170 Solutions**

**Thread Play **

Take a 15 cm long thread. Make different shapes by joining their ends on this sheet.

**(a) Which shape has the biggest area? How much? What is the perimeter of this shape?**

**(b) Which shape has the smallest area? How much? What is the perimeter of this shape?**

**Also make a triangle, a square, a rectangle and a circle.**

**Find which shape has the biggest area and which has the smallest.**

Ans. (a) For shape A

Area = (9 + 2 x 1/2 + 7)

= 17 square cm

Number of complete squares = 9 Number of more than half squares = 7 Number of half squares = 2

Area of shape B

= (2 + 2 x 1/2 + 7)

= 10 square cm

Area of shape C

= (9 + 1 x 1/2 + 7)

= 16 (1/2) square cm

So, shape A has the biggest area

The perimeter of this shape is 15 cm (because the length of the thread is 15 cm)

(b) Shape B has the smallest area.

The perimeter of this shape = is 15 cm (because the length of the thread is 15 cm) Area of the triangle = (6 + 0 x 1/2 + 6)

= 12 square cm

Area of square = (9 + 0 x 1/2 + 7)

= 16 square cm

Area of rectangle = (10 + 5 x 1/2 + 0)

= 12 square cm

**Page No. 171 Solutions**

**Save the Birds:**

**There are two beautiful lakes near a village. People come for boating and picnics on both lakes. The village Panchayat is worried that with the noise of the boats, the birds will stop coming. The Panchayat wants motor boats in only one lake. The other lake will be saved for the birds to make their nests.**

**a. How many cm is the length of the boundary of Lake A in the drawing? __________ (use thread to find out)**

Ans. The length of the boundary of Lake A in the drawing is 33 cm.

**b. What is the length of the boundary of Lake B in the drawing?**

Ans. The length of the boundary of Lake B in the drawing is 26 cm.

**c. How many kilometres long is the actual boundary of Lake A?**

Ans. Since 1 cm on drawing = 1 km on the ground

33 cm represents 33 × 1 km i.e. 33 km.

**d. How many kilometres long is the actual boundary of Lake B?**

Ans. The actual boundary of Lake B is 26 km.

**e. A longer boundary around the lake will help more birds to lay their eggs. So which lake should be kept for birds? Which lake should be used for boats?**

Ans. Lake A is longer than that of Lake B. So, it may be kept for birds. The lake B should be used for boats.

**f. Find the area of lake B on the drawing in square cm. What is the actual area in square km?**

Ans. Lake B contains 15 complete squares, 3 half squares, 8 more than half squares and 2 less than half squares. Neglecting less than half squares and considering more than half squares the approximate area of lake B on drawing.

= (15 + 3 × 1/2 + 8)

= (15 + 3/2 + 8)

= (30 + 3 + 16)/2 = 49/2

= 24 ½ square cm.

Therefore, the area of Lake B = 24 ½ square cm.

**Page No. 172 Solutions**

**King’s Story:**

The king was very happy with the carpenters. Balbir and Kuldeep. They had made a very big and beautiful bed for him. So as gifts the king wanted to give some land to Balbir, and some gold to Kuldeep.

Balbir was happy. He took 100 meters of wire and tried to make different rectangles.

He made a 10 m × 40 m rectangle.

Its area was 400 square meters.

So, he next made a 30 m × 20 m rectangle.

**What is its area? Is it more than the first rectangle?**

Ans. The area of this rectangle = 30 m 20 m = 600 square m.

Yes, its area is more than that of the first rectangle.

**What other rectangle can he make with 100 metres of wire? Discuss which of these rectangles will have the biggest area.**

Many other rectangles can be made with 100 metres of wire. For Example,

5m 45m having an area of 225 square metres.

15m 35m having an area of 525 square metres.

20 m 30m having an area of 600 square metres.

25m 25 m having an area of 625 square metres.

The rectangle having length = breadth i.e. a square will have the biggest area.

Balbir’s wife asked him to make a circle with a wire. She knew it had an area of 800 square meters.

**Page No. 173 – 174 Solutions**

**Why did Balbir not choose a rectangle? Explain.**

Ans. Because his wife knew that a circle made with this wire would have an area of 800 square metres.

So, Kuldeep also tried many different ways to make a boundary for 800 square metres of land.

**He made rectangles A, B, and C of different sizes. Find out the length of the boundary of each. How much gold wire will he get for these rectangles?**

**(a) 40 m x 20 m**

Perimeter = 2(length + breadth) = 2 (40 + 20) = 2 x 60 = 120 m

**(b) 80 m x 10 m**

Perimeter = 2(length + breadth) = 2(80 + 10) = 2 x 90 = 180 m

**(c) 800 m x 1 m**

Perimeter = 2(length + breadth) = 2 (800 + 1) = 2 x 801 = 1602 m

But then Kuldeep made an even longer rectangle See how long!

**(d) 8000 m x 0.1 m**

Perimeter = 2(length + breadth) = 2(8000 + 0.1) = 2 × 8000.1 = 16000.2 m

**Now do you understand why the king fainted!!**

Ans. The king fainted because he could not arrange so much gold.

**Can you make a rectangle with a still longer boundary? I made a rectangle 1 cm wide and 80000 m long. Imagine how long that boundary will be!!! With that much gold wire, I can become a king!**

Ans. Perimeter = 2(length + breadth)

2(80000 m + 1 cm)

= 2(80000 m + 1/100 m)

= 2(80000 m + 0.01 m)

= 2 × 80000.01 m = 16000.02 m

**Page No. 175 Solutions**

**Now Lets Us Do These**

**Find the area of the rectangle with sides 30 cm and 50 cm.**

Ans. Area of the rectangle = Length × Width

= 50 cm × 30 cm = 1500 sq. cm.

**If the area of a square ‘A’ is 1600 sq. cm and if the area of a square B is 40 sq. cm then find the number of squares of type B obtained from square A.**

Ans. Number of squares of type B = Area of square A/Area of square B

= 1600/40 = 40

**If the area of a rectangle is 250 sq. cm and the length of a rectangle is 25 cm then find the width of the rectangle.**

Ans. Width of the rectangle = Area of rectangle/Length of the rectangle

= 250/25 = 10 cm.

**If the dimensions of a rectangle are 20 cm and 10 cm, then find the perimeter of a rectangle.**

Ans. Length of the rectangle = 20 cm.

Breadth of the rectangle = 10 cm.

The perimeter of the rectangle = 2(length + breadth)

= 2(20 + 10) = 2 × 30 = 60 cm.

**If the perimeter of a rectangle is 60 cm and it is 10 cm wide. Find its length.**

Ans. Length of a rectangle = ?

The width of the rectangle = 10 cm.

The perimeter of the rectangle = 60 cm.

So, 2(Length + Breadth) = 60 cm

Length + Breadth = 30 cm.

So, Length + 10 cm = 30 cm.

Length = 30 cm – 10 cm = 20 cm.

**From the figure find the area of region B.**

Ans. Length of region B = 40 – (4 cm + 4 cm) = 32 cm

Width of region B = 30 – (2 cm + 2 cm) = 26 cm.

Area of the region B = 32 cm × 26 cm = 832 sq. cm.

**In a rectangular plot of land:**

**A: Plot of land **

**B: Wetland **

**Find the area of B when the area of A = 200 Sq. mts. **

Ans. Area of whole rectangle = Length × Width

= 40 cm × 10 cm

= 400 sq. cm

Area of A = 200 sq. cm.

So, Area of B = 400 sq. cm – 200 sq.cm = 200 sq. cm

That’s all about

**Area and its Boundary ****Class 5 JKBOSE Solutions****. **Hope it helped. Do share your views about this post in the comment section below.

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